Question

Solve This Integral

Answer 1

What integral?

Answer 2

\[\int\limits_{0}^{\infty} \frac{ x dx}{ e^{x} -1 }\]

Answer 3

Please show the steps.
I guess i should use residue theorem but i don't know how

Answer 4

What have you done so far?

Answer 5

I know that if x = 0 the denominator is 0, so one residue is Res1 = x/(d/dx (e^x - 1)) = x/e^x evaluated at x = 0 whitch gives 0, but i'm not sure if it's right

Answer 6

This is indefinite integral.
First, instead of infinity, put t in.
\[\int\limits_{0}^{t}\frac{ x }{ e^x-1 }dx\]

Answer 7

Then solve it in terms of t

Answer 8

Ok sure, but how i solve it?

Answer 9

i must solve it using residue theorem

Answer 10

\[\int_0^\infty \frac{x}{e^x-1}~dx\]
Recall the power series for \(e^x\):
\[e^x=1+x+\frac{x^2}{2}+\cdots\]
Then you have
\[\frac{x}{e^x-1}=\frac{x}{1+x+\frac{x^2}{2}+\cdots-1}=\frac{x}{x+\frac{x^2}{2}+\cdots}=\frac{1}{1+\frac{x}{2}+\cdots}\]
Since \(z=0\) is an isolated singular point and \(\displaystyle\lim_{z\to0}\frac{z}{e^z-1}=1\not=\infty\), \(z=0\) is a removable singularity, so the residue is 0.

As for computing the integral... I think you must take the principal value. WolframAlpha doesn't seem to be able to compute without doing just that.

Answer 11

Ok, now that i have the value of the residue, how can i express the solution? i know, from WolframAlpha, the result is pi²/6

Answer 12

here is another way to evaluate the integral,